Probability and Sampling Distributions Randomness and Probability Models PBS Chapters 4. 1 and 4. 2 © 2009 W. H. Freeman and Company
Objectives (PBS Chapters 4. 1 and 4. 2) Randomness and Probability Models p Randomness and probability p Probability rules p Assigning probabilities: finite number of outcomes p Assigning probabilities: intervals of outcomes p Normal probability models
Randomness and probability A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions. The probability of any outcome of a random phenomenon can be defined as the proportion of times the outcome would occur in a very long series of repetitions.
Coin toss The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i. e. , the outcome of a new coin flip is not influenced by the result of the previous flip). The probability of heads is 0. 5 = the proportion of times you get heads in many repeated trials. First series of tosses Second series
Two events are independent if the probability that one event occurs on any given trial of an experiment is not affected or changed by the occurrence of the other event. When are trials not independent? Imagine that these coins were spread out so that half were heads up and half were tails up. Close your eyes and pick one. The probability of it being heads is 0. 5. However, if you don’t put it back in the pile, the probability of picking up another coin and having it be heads is now less than 0. 5. The trials are independent only when you put the coin back each time. It is called sampling with replacement.
Probability models describe mathematically the outcome of random processes and consist of two parts: 1) S = Sample Space: This is a set, or list, of all possible outcomes of a random process. An event is a subset of the sample space. 2) A probability for each possible event in the sample space S. Example: Probability Model for a Coin Toss: S = {Head, Tail} Probability of heads = 0. 5 Probability of tails = 0. 5
Sample spaces It’s the question that determines the sample space. H - HHH A. A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)? H M - HHM H M M … H - HMH M - HMM … B. A basketball player shoots three free throws. What is the number of baskets made? S = { HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM } S = { 0, 1, 2, 3 } Note: 8 elements, 23
Probability rules 1) Probabilities range from 0 (no chance of the event) to 1 (the event has to happen). For any event A, 0 ≤ P(A) ≤ 1 Coin Toss Example: S = {Head, Tail} Probability of heads = 0. 5 Probability of tails = 0. 5 Probability of getting a Head = 0. 5 We write this as: P(Head) = 0. 5 P(neither Head nor Tail) = 0 P(getting either a Head or a Tail) = 1 2) Because some outcome must occur on every trial, the sum of the probabilities Coin toss: S = {Head, Tail} for all possible outcomes (the sample space) must be exactly 1. P(head) + P(tail) = 0. 5 + 0. 5 =1 P(sample space) = 1
Probability rules (cont ) d Coin Toss Example: S = {Head, Tail} Probability of heads = 0. 5 Probability of tails = 0. 5 3) The complement of any event A is the event that A does not occur, written as Ac. The complement rule states that the probability of an event not occurring is 1 minus the probability that is does occur. P(not A) = P(Ac) = 1 − P(A) Tailc = not Tail = Head P(Tailc) = 1 − P(Head) = 0. 5 Venn diagram: Sample space made up of an event A and its complementary Ac, i. e. , everything that is not A.
Probability rules (cont ) d Venn diagrams: A and B disjoint 4) Two events A and B are disjoint if they have no outcomes in common and can never happen together. The probability that A or B occurs is then the sum of their individual probabilities. P(A or B) = “P(A U B)” = P(A) + P(B) This is the addition rule for disjoint events. A and B not disjoint Example: If you flip two coins, and the first flip does not affect the second flip: S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0. 25. The probability that you obtain “only heads or only tails” is: P(HH or TT) = P(HH) + P(TT) = 0. 25 + 0. 25 = 0. 50
Assigning Probabilities: finite number of outcomes Finite sample spaces deal with discrete data — data that can only take on a limited number of values. These values are often integers or whole numbers. Throwing a die: S = {1, 2, 3, 4, 5, 6} The individual outcomes of a random phenomenon are always disjoint. The probability of any event is the sum of the probabilities of the outcomes making up the event (addition rule).
M&M candies If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here: Color Probability Brown Red Yellow Green Orange Blue 0. 3 0. 2 0. 1 ? What is the probability that an M&M chosen at random is blue? S = {brown, red, yellow, green, orange, blue} P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)] = 1 – [0. 3 + 0. 2 + 0. 1] = 0. 1 What is the probability that a random M&M is any of red, yellow, or orange? P(red or yellow or orange) = P(red) + P(yellow) + P(orange) = 0. 2 + 0. 1 = 0. 5
Probabilities: equally likely outcomes We can assign probabilities either: p empirically from our knowledge of numerous similar past events p Mendel discovered the probabilities of inheritance of a given trait from experiments on peas without knowing about genes or DNA. p or theoretically from our understanding the phenomenon and symmetries in the problem p A 6 -sided fair die: each side has the same chance of turning up p Genetic laws of inheritance based on meiosis process If a random phenomenon has k equally likely possible outcomes, then each individual outcome has probability 1/k. And, for any event A:
Dice You toss two dice. What is the probability of the outcomes summing to 5? This is S: {(1, 1), (1, 2), (1, 3), ……etc. } There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36. P(the roll of two dice sums to 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4 / 36 = 0. 111
The gambling industry relies on probability distributions to calculate the odds of winning. The rewards are then fixed precisely so that, on average, players lose and the house wins. The industry is very tough on so-called “cheaters” because their probability to win exceeds that of the house. Remember that it is a business, and therefore it has to be profitable.
p A couple wants three children. What are the arrangements of boys (B) and girls (G)? Genetics tell us that the probability that a baby is a boy or a girl is the same, 0. 5. Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} All eight outcomes in the sample space are equally likely. The probability of each is thus 1/8. Each birth is independent of the next, so we can use the multiplication rule. Example: P(BBB) = P(B)* P(B) = (1/2)*(1/2) = 1/8 A couple wants three children. What are the numbers of girls (X) they could have? The same genetic laws apply. We can use the probabilities above and the addition rule for disjoint events to calculate the probabilities for X. Sample space: {0, 1, 2, 3} P(X = 0) = P(BBB) = 1/8 P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 3/8 …
Assigning Probabilities: intervals of outcomes p A sample space may contain all numbers within a range. p For continuous outcomes, the probability model is a density curve. p Area under the entire density curve is equal to 1. p Probability model assigns probabilities as areas under the density curve.
Assigning Probabilities: intervals of outcomes Software random number generators may use S = {all numbers between 0 and 1} All possible outcomes are equally likely. The results of many trials are represented by the uniform density curve. Probabilities are computed as areas P(0. 3 X 0. 7) = 0. 4 Similarly, P(X < 0. 5 or X > 0. 8) = 0. 5 +0. 2 = 0. 7
Normal probability models p Normal distributions are probability models. q The scores of students on the ACT college entrance examination in a recent year had the normal distribution with mean =18. 6 and standard deviation = 5. 9. What is the probability that a randomly chosen student scores 21 or higher? q The calculation is the same as those we did in Chapter 1. Only the language of probability is new.
Probability and Sampling Distributions Random variables PBS Chapter 4. 3 © 2009 W. H. Freeman and Company
Objectives (PBS Chapter 4. 3) Random Variables p Random variables p Probability distributions p Mean of a random variable p Variance of a random variable p Rules for means and variances
Random variable A random variable is a variable whose value is a numerical outcome of a random phenomenon. A basketball player shoots three free throws. We define the random variable X as the number of baskets successfully made. A discrete random variable X has a finite number of possible values. A continuous random variable X takes all values in an interval.
Probability distributions p The probability distribution of a random variable X tells us what values X can take and how to assign probabilities to those values. p Because of the differences in the nature of sample spaces for discrete and continuous sample random variables, we describe probability distributions for the two types of random variables separately.
The probability distribution of a discrete random variable X lists the values and their probabilities: The probabilities pi must add up to 1. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made. H - HHH H Value of X M - HHM H M M … 0 1 2 3 Probability 1/8 3/8 1/8 MMM HMM MHM MMH HHM HMH MHH H - HMH M - HMM …
The probability of any event is the sum of the probabilities pi of the values of X that make up the event. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made. What is the probability that the player Value of X 0 1 2 3 successfully makes at least two Probability 1/8 3/8 1/8 MMM HMM MHM MMH HHM HMH MHH HHH baskets (“at least two” means “two or more”)? P(X≥ 2) = P(X=2) + P(X=3) = 3/8 + 1/8 = 1/2 What is the probability that the player successfully makes fewer than three baskets? P(X<3) = P(X=0) + P(X=1) + P(X=2) = 1/8 + 3/8 = 7/8 or P(X<3) = 1 – P(X=3) = 1 – 1/8 = 7/8
Continuous Probability Distributions A continuous random variable X takes all values in an interval. Example: There is an infinity of numbers between 0 and 1 (e. g. , 0. 001, 0. 4, 0. 0063876). The probability distribution of a continuous random variable is described by a density curve. The probability of any event is the area under the density curve for the values of X that make up the event. This is a uniform density curve for the variable X. The probability that X falls between 0. 3 and 0. 7 is the area under the density curve for that interval: P(0. 3 ≤ X ≤ 0. 7) = (0. 7 – 0. 3)*1 = 0. 4 X
Intervals All Continuous probability distributions assign probability 0 to every individual outcome. Only intervals can have a positive probability, represented by the area under the density curve for that interval. The probability of a single event is zero: P(X=1) = (1 – 1)*1 = 0 Height = 1 The probability of an interval is the same whether boundary values are included or excluded: P(0 ≤ X ≤ 0. 5) = (0. 5 – 0)*1 = 0. 5 P(0 < X < 0. 5) = (0. 5 – 0)*1 = 0. 5 X P(0 ≤ X < 0. 5) = (0. 5 – 0)*1 = 0. 5 P(X < 0. 5 or X > 0. 8) = P(X < 0. 5) + P(X > 0. 8) = 1 – P(0. 5 < X < 0. 8) = 0. 7
Continuous random variable and population distribution The shaded area under a density curve shows the proportion, or %, of individuals in a population with values of X between x 1 and x 2. Because the probability of drawing one individual at random depends on the frequency of this type of individual in the population, the probability is also the shaded area under the curve. % individuals with X such that x 1 < X < x 2
Normal probability distributions The probability distribution of many random variables is a normal distribution. It shows what values the random variable can take and is used to assign probabilities to those values. Example: Probability distribution of women’s heights. Here since we chose a woman randomly, her height, X, is a random variable. To calculate probabilities with the normal distribution, we will standardize the random variable (z score) and use Table A.
What is the probability, if we pick one woman at random, that her height will be some value X? For instance, between 68 and 70 inches P(68 < X < 70)? Because the woman is selected at random, X is a random variable. N(µ, ) = N(64. 5, 2. 5) As before, we calculate the zscores for 68 and 70. For x = 68", For x = 70", 0. 9192 0. 9861 The area under the curve for the interval [68" to 70"] is 0. 9861 − 0. 9192 = 0. 0669. Thus, the probability that a randomly chosen woman falls into this range is 6. 69%. P(68 < X < 70) = 6. 69%
Mean of a random variable The mean of a set of observations is their arithmetic average. The mean µ of a random variable X is a weighted average of the possible values of X, reflecting the fact that all outcomes might not be equally likely. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made (“H”). MMM HMM MHM MMH HHM HMH MHH Value of X HHH 0 1 2 3 Probability 1/8 3/8 1/8 The mean of a random variable X is also called expected value of X.
Mean of a discrete random variable For a discrete random variable X with probability distribution the mean µ of X is found by multiplying each possible value of X by its probability, and then adding the products. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made. Value of X 0 1 2 3 Probability 1/8 3/8 1/8 The mean µ of X is µ = (0*1/8) + (1*3/8) + (2*3/8) + (3*1/8) = 12/8 = 3/2 = 1. 5
Variance of a random variable The variance and the standard deviation are the measures of spread that accompany the choice of the mean to measure center. The variance σ2 X of a random variable is a weighted average of the squared deviations (X − µX)2 of the variable X from its mean µX. Each outcome is weighted by its probability in order to take into account outcomes that are not equally likely. The larger the variance of X, the more scattered the values of X on average. The positive square root of the variance gives the standard deviation σ of X.
Variance of a discrete random variable For a discrete random variable X with probability distribution and mean µX, the variance σ2 of X is found by multiplying each squared deviation of X by its probability and then adding all the products. A basketball player shoots three free throws. The random variable X is the number of baskets successfully made. µX = 1. 5. The variance σ2 of X is Value of X 0 1 2 3 Probability 1/8 3/8 1/8 σ2 = 1/8*(0− 1. 5)2 + 3/8*(1− 1. 5)2 + 3/8*(2− 1. 5)2 + 1/8*(3− 1. 5)2 = 2*(1/8*9/4) + 2*(3/8*1/4) = 24/32 = 3/4 =. 75
Rules for means and variances If X is a random variable and a and b are fixed numbers, then µa+b. X = a + bµX σ2 a+b. X = b 2σ2 X If X and Y are two independent random variables, then µX+Y = µX + µY σ2 X+Y = σ2 X + σ2 Y σ2 X-Y = σ2 X + σ2 Y If X and Y are NOT independent but have correlation ρ, then µX+Y = µX + µY σ2 X+Y = σ2 X + σ2 Y + 2ρσXσY σ2 X-Y = σ2 X + σ2 Y - 2ρσXσY
Investment You invest 20% of your funds in Treasury bills and 80% in an “index fund” that represents all U. S. common stocks. Your rate of return over time is proportional to that of the T-bills (X) and of the index fund (Y), such that R = 0. 2 X + 0. 8 Y. Based on annual returns between 1950 and 2003: p Annual return on T-bills µX = 5. 0% σX = 2. 9% p Annual return on stocks µY = 13. 2% σY = 17. 6% p Correlation between X and Yρ = − 0. 11 µR = 0. 2µX + 0. 8µY = (0. 2*5) + (0. 8*13. 2) = 11. 56% σ2 R = σ20. 2 X + σ20. 8 Y + 2ρσ0. 2 Xσ0. 8 Y = 0. 2*2σ2 X + 0. 8*2σ2 Y + 2ρ*0. 2*σX*0. 8*σY = (0. 2)2(2. 9)2 + (0. 8)2(17. 6)2 + (2)(− 0. 11)(0. 2*2. 9)(0. 8*17. 6) = 196. 786 σR = √ 196. 786 = 14. 03% The portfolio has a smaller mean return than an all-stock portfolio, but it is also less risky.
Probability and Sampling Distributions The Sampling Distribution of a Sample Mean PBS Chapter 4. 4 © 2009 W. H. Freeman and Company
Objectives (PBS Chapter 4. 4) Sampling distribution of a sample mean p Law of large numbers p Sampling distributions p The mean and standard deviation of p The central limit theorem
Law of large numbers As the number of randomly drawn observations in a sample increases, the mean of the sample gets closer and closer to the population mean m. This is the law of large numbers. It is valid for any population. Note: We often intuitively expect predictability over a few random observations, but it is wrong. The law of large numbers only applies to really large numbers.
Reminder: What is a sampling distribution? The sampling distribution of a statistic is the distribution of all possible values taken by the statistic when all possible samples of a fixed size n are taken from the population. It is a theoretical idea — we do not actually build it. The sampling distribution of a statistic is the probability distribution of that statistic.
Sampling distribution of sample mean We take many random samples of a given size n from a population with mean m and standard deviation s. Some sample means will be above the population mean m and some will be below, making up the sampling distribution. Sampling distribution of “x bar” Histogram of some sample averages
For any population with mean m and standard deviation s : p The mean of the sampling distribution is equal to the population mean m. The standard deviation of the sampling distribution is s/√n, where n is the sample size. p Sampling distribution of x bar s/√n m
Mean and standard deviation of sample mean Mean of a sampling distribution of There is no tendency for a sample mean to fall systematically above or below m, even if the distribution of the raw data is skewed. Thus, the mean of the sampling distribution is an unbiased estimate of the population mean m — it will be “correct on average” in many samples. Standard deviation of a sampling distribution of The standard deviation of the sampling distribution is smaller than the standard deviation of the population by a factor of √n. Averages are less variable than individual observations. Also, the results of large samples are less variable than the results of small samples.
For normally distributed populations When a variable in a population is normally distributed, the sampling distribution of the sample mean for all possible samples of size n is also normally distributed. Sampling distribution If the population is N(m, s) then the sample means distribution is N(m, s/√n). Population
The central limit theorem Central Limit Theorem: When randomly sampling from any population with mean m and standard deviation s, when n is large enough, the sampling distribution of x bar is approximately normal: ~ N(m, s/√n). Population with strongly skewed distribution Sampling distribution of for n = 10 observations Sampling distribution of for n = 2 observations Sampling distribution of for n = 25 observations
IQ scores: population vs. sample In a large population of adults, the mean IQ is 112 with standard deviation 20. Suppose 200 adults are randomly selected for a market research campaign. p. The distribution of the sample mean IQ is: A) Exactly normal, mean 112, standard deviation 20 B) Approximately normal, mean 112, standard deviation 20 C) Approximately normal, mean 112 , standard deviation 1. 414 D) Approximately normal, mean 112, standard deviation 0. 1 C) Approximately normal, mean 112 , standard deviation 1. 414
Application Hypokalemia is diagnosed when blood potassium levels are low, below 3. 5 m. Eq/dl. Let’s assume that we know a patient whose measured potassium levels vary daily according to a normal distribution N(m = 3. 8, s = 0. 2). If only one measurement is made, what is the probability that this patient will be misdiagnosed hypokalemic? z = − 1. 5, P(z < − 1. 5) = 0. 0668 ≈ 7% If instead measurements are taken on 4 separate days, what is the probability of such a misdiagnosis? z = − 3, P(z < − 1. 5) = 0. 0013 ≈ 0. 1% Note: Make sure to standardize (z) using the standard deviation for the sampling distribution.
Income distribution Let’s consider the very large database of individual incomes from the Bureau of Labor Statistics as our population. It is strongly right skewed. p We take 1000 SRSs of 100 incomes, calculate the sample mean for each, and make a histogram of these 1000 means. p We also take 1000 SRSs of 25 incomes, calculate the sample mean for each, and make a histogram of these 1000 means. Which histogram corresponds to the samples of size 100? 25?
How large a sample size? It depends on the population distribution. More observations are required if the population distribution is far from normal. p A sample size of 25 is generally enough to obtain a normal sampling distribution from a strong skewness or even mild outliers. p A sample size of 40 will typically be good enough to overcome extreme skewness and outliers. In many cases, n = 25 isn’t a huge sample. Thus, even for strange population distributions we can assume a normal sampling distribution of the mean and work with it to solve problems.
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